What is the Acceleration of a Tennis Ball After Being Served?

AI Thread Summary
To determine the acceleration of a tennis ball served at 49 m/s over a distance of 45 cm, one can use kinematic equations. The relevant equations are v = at and s = (1/2)at², where v is final velocity, a is acceleration, and s is distance. By substituting the known values into these equations, the acceleration can be calculated without needing time. Once acceleration is found, the net force can be calculated using Fnet = ma. The discussion emphasizes the importance of using kinematic formulas correctly to solve for acceleration and net force.
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When a 58 g tennis ball is served, it accelerates from rest to a constant speed of 49 m/s. The impact with the racket gives the ball a constant acceleration over a distance of 45 cm. What is the magnitude of the net force acting on the ball?

I tried using Fnet=ma, but acceleration isn't given. So I tried to find acceleration by dividing the velocity by time, but time isn't given. Then I tried to use a kinematic formula without time in it to solve for acceleration. When I got my "answer", it wasn't right. What in the world am I doing wrong?
 
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How much work has the force done on the ball?
 
For any acceleration, starting from rest, v= at and s= (1/2)at2.

Here you are told that v= 49 m/s and s= 45 m/s2. You can solve those two equations for a and t. (Of course, you only need a.)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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